Question

Solve it step by step.
Don't give wrong answer.

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Answer 1

10.⤵️⤵️⤵️

3a-2b+5c=5

Cubing both sides,

{(3a-2b)+5c}³=5³

or, (3a-2b)³+3.(3a-2b)².5c+3(3a-2b).(5c)²+125c³=125

or, 27a³-3.9a².2b+3.3a.4b²-8b³+3.(9a²-2.3a.2b+4b²).5c+(9a-6b)(25c²)+  125c³=125

or, 27a³-54a²b+36ab²-8b³+(27a²-36ab+12b²).5c+225ac²-150bc²+125c³=125

or, 27a³-54a²b+36ab²-8b³+135a²c-180abc+60b²c+225ac²-150bc²+125c³=125

or, 27a³-8b³+125c³-54a²b-90abc+135a²c+36ab²+60b²c-90abc-90abc-150bc²+225ac²+90abc=125

or, 27a³-8b³+125c³-9a(6ab+10bc-15ac)+6b(6ab+10bc-15ac)-15c(6ab+10bc-15ac)+90abc=125

or, 27a³+125c³-8b³+90abc-9a×14+6b×14-15c×14=125 

[∵, 6ab+10bc-15ac=14]

or, 27a³+125c³+90abc-8b³-126a+84b-210c=125

or, 27a³+125c³+90abc-8b³-42(3a-2b+5c)=125

or, 27a³+125c³+90abc-8b³-42×5=125 [∵, 3a-2b+5c=5]

or, 27a³+125c³+90abc-8b³-210=125

or, 27a³+125c³+90abc-8b³=125+210

or, 27a³+125c³+90abc-8b³=335

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11.⤵️⤵️⤵️

We know that,

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

If a+b+c=0 then,

a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)

let a=(x-a)

b=(x-b)

and c=(x-c)

a+b+c=x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0....(as a+b+c=3x)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0

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12. ⤵️⤵️⤵️

Since a+b+c = 0

a^3 + b^3 +c^3 + 3abc

a^3/abc + b^3/abc + c^3/abc = 3

a^2/bc + b^2/ac + c^2/ab = 3

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Answer 2

Answer:

Step-by-step explanation:

(3a-2b+5c)^2=5^2

9a^2+4b^2+25c^2-2(6ab+10bc-15ca)=25

=>9a^2+4b^2+25c^2-2(14)=25

=>9a^2+4b^2+25c^2=25-28=-3

gIVen:27a^3+125c^3+90abc-8b^3

=(3a)^3+(5c)^3+(-2b)^3-(3*3a*5c*-2b)

=(3a+5c-2b)(9a^2+4b^2+25c^2-(3a*5c+5c*-2b+(-2b*3a))

=5(-3-2(15ac-6ab-10bc))

=5(-3+2(14))

=5*25

=125